 Solution

Use signature $(+,-,-,-)$, which is required for the stated tensor convention. Then $U_b=\gamma (1,-\mathbf{v})$. The time component is $F^{0b}{U}_b=\gamma \mathbf{E}\cdot \mathbf{v}$, while $F^{ij}=-{ϵ}_{ijk}{B}_k$ gives the spatial components $\gamma (\mathbf{E}+\mathbf{v}\times \mathbf{B})$. Thus

The scalar ${\rho }_{\ast }=J^b{U}_b$ is the charge density in the medium's rest frame. In that frame $U=(1,0)$, so $J-{\rho }_{\ast }U=(0,\mathbf{j})$ and $FU=(0,\mathbf{E})$. The rest-frame Ohm's law is precisely their proportionality with rest conductivity $\sigma$. Both sides are four-vectors, so its covariant form isThe time and spatial components respectively giveEliminate ${\rho }_{\ast }$ to obtainIf the rest-frame charge density vanishes, set ${\rho }_{\ast }=0$, not necessarily $\rho =0$ in the lab frame. ThenThe induced lab charge is the term that cancels the longitudinal correction in the general formula.